If `3 + (1)/(4) (3 + d) + (1)/(4^(2)) (3 + 2d)+…oo = 8`, then the value of d is

Options

A. 1
B. 5
C. 9
D. none

Answer

To solve the equation \( 3 + \frac{1}{4}(3 + d) + \frac{1}{4^2}(3 + 2d) + \ldots = 8 \), we can follow these steps:

### Step 1: Define the series
Let \( S \) be the sum of the series:
\[ S = 3 + \frac{1}{4}(3 + d) + \frac{1}{4^2}(3 + 2d) + \ldots \]

### Step 2: Factor out common terms
We can factor out \( \frac{1}{4} \) from the series starting from the second term:
\[ S = 3 + \frac{1}{4} \left( (3 + d) + \frac{1}{4}(3 + 2d) + \ldots \right) \]

### Step 3: Rewrite the series
Let’s denote the series starting from the second term as \( T \):
\[ T = (3 + d) + \frac{1}{4}(3 + 2d) + \frac{1}{4^2}(3 + 3d) + \ldots \]
Thus, we can write:
\[ S = 3 + \frac{1}{4}T \]

### Step 4: Express \( T \) in terms of \( S \)
Now, we can express \( T \) in terms of \( S \):
\[ T = (3 + d) + \frac{1}{4}(3 + d) + \frac{1}{4^2}(3 + d) + \ldots \]
This is a geometric series with the first term \( (3 + d) \) and a common ratio of \( \frac{1}{4} \):
\[ T = (3 + d) \left( 1 + \frac{1}{4} + \frac{1}{4^2} + \ldots \right) \]
The sum of the infinite geometric series is given by \( \frac{a}{1 - r} \):
\[ T = (3 + d) \cdot \frac{1}{1 - \frac{1}{4}} = (3 + d) \cdot \frac{4}{3} \]

### Step 5: Substitute \( T \) back into \( S \)
Now substitute \( T \) back into the equation for \( S \):
\[ S = 3 + \frac{1}{4} \cdot \left( (3 + d) \cdot \frac{4}{3} \right) \]
\[ S = 3 + \frac{1}{3}(3 + d) \]
\[ S = 3 + 1 + \frac{d}{3} \]
\[ S = 4 + \frac{d}{3} \]

### Step 6: Set the equation equal to 8
Now we set \( S \) equal to 8:
\[ 4 + \frac{d}{3} = 8 \]

### Step 7: Solve for \( d \)
Subtract 4 from both sides:
\[ \frac{d}{3} = 4 \]
Multiply both sides by 3:
\[ d = 12 \]

### Final Answer
The value of \( d \) is \( 12 \).